Question: Question 2, 6.6.1 Part 1 of 4 In 2012, the population of a city was 6.07 million. The exponential…

Question 2, 6.6.1 Part 1 of 4 In 2012, the population of a city was 6.07 million. The exponential growth rate was 1.84% per year. a) Find the exponential growth function. b) Estimate the population of the city in 2018. c) When will the population of the city be 10 million? d) Find the doubling time. a) The exponential growth function is P(t) = _, where t is in terms of the number of years since 2012 and P(t) is the population in millions. (Type exponential notation with positive exponents. Do not simplify. Use integers or decimals for any numbers in the equation.)

Solution

In 2012, the population of a city was 6.07 million with an exponential growth rate of 1.84% per year. a) Find the exponential growth function. The exponential growth function is given by: \[ P(t) = P_0 \cdot (1 + r)^t \] where: - \( P_0 = 6.07 \) million (initial population in 2012), - \( r = 0.0184 \) (growth rate), - \( t \) is the time in years since 2012. Substitute the values: \[ P(t) = 6.07 \cdot (1 + 0.0184)^t \] \[ P(t) = 6.07 \cdot (1.0184)^t \] b) Estimate the population of the city in 2018. For 2018, \( t = 2018 - 2012 = 6 \). Substitute \( t = 6 \) into the exponential function: \[ P(6) = 6.07 \cdot (1.0184)^6 \] Calculate \( P(6) \): \[ P(6) \approx 6.07 \cdot 1.116 \] \[ P(6) \approx 6.77 \text{ million} \] c) When will the population of the city be 10 million? Set \( P(t) = 10 \) and solve for \( t \): \[ 10 = 6.07 \cdot (1.0184)^t \] Divide both sides by 6.07: \[ 1.647 \approx (1.0184)^t \] Take the natural logarithm of both sides: \[ \ln(1.647) \approx t \cdot \ln(1.0184) \] Solve for \( t \): \[ t \approx \frac{\ln(1.647)}{\ln(1.0184)} \] \[ t \approx 26.3 \text{ years} \] Therefore, the population will reach 10 million approximately 26 years after 2012, which would be around 2038. d) Find the doubling time. The doubling time \( T \) can be found using the rule: \[ T = \frac{\ln(2)}{\ln(1 + r)} \] Substitute the growth rate: \[ T = \frac{\ln(2)}{\ln(1.0184)} \] Calculate \( T \): \[ T \approx \frac{0.693}{0.0183} \] \[ T \approx 37.9 \text{ years} \] The doubling time is approximately 38 years.